Is an experiment planned to discern determinism and randomness in QM

In summary: Since the measurements can be arranged to be space-like, this experiment proves that the measurements CANNOT be random, and have to be predetermined.
  • #106
PeroK said:
The state of B cannot change, because B had no state before measurement. The measurement breaks the entanglement after which A and B have single-particle states.
Before the A measurement B "had no state"
After the A measurement B has a "single-particle" state.

This is, by any definition a change. The A measurement changes B. The Bohmian in you starts to emerge.

PeroK said:
It's clear you do not understand this or its implications for EPR.
Well, you seem not to comprehend the meaning of the word "change".
 
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  • #107
Demystifier said:
To me, all this is just an explanation why QBism is indeed a local interpretation.
Using your definition of locality, everything is local. You simply define-away the concept. If an instant teleporter is local, everything is local. The problem with this definition is that some "local" objects are incompatible with relativity. Just defining a teleporter "local" does not make it compatible with relativity, so the fact that QBism is local according to your definition does not resolve the conflict. QBism remains incompatible with relativity.
 
  • #108
AndreiB said:
Using your definition of locality, everything is local. You simply define-away the concept. If an instant teleporter is local, everything is local.
Not mine, but those of QBists. I'm not their fan, I'm just trying to explain what they say. But yes, according to them everything is local. Even if entanglement could be used for instantaneous communication between two agents, it would still be local for them because each agent is a local being.

AndreiB said:
The problem with this definition is that some "local" objects are incompatible with relativity. Just defining a teleporter "local" does not make it compatible with relativity, so the fact that QBism is local according to your definition does not resolve the conflict. QBism remains incompatible with relativity.
What physical mechanism do you have in mind for a teleporter? The only one that occurs to me is a wormhole, which is local. Besides teleporter, do you suggest any other ways how QBism conflicts relativity?
 
  • #109
AndreiB said:
Well, you seem not to comprehend the meaning of the word "change".
The two-particle system changes to two independent particles. The discipline of orthodox QM is to focus on states; not the realism of always independent particles.

In a deeper sense there is no A and B before measurement. That there is no classical analogue is one reason you need to think very differently about QM. If you approach QM as you have done with a rigidly realistic mindset, then it's not surprising it causes consternation.

This is at the root of EPR: the realist view versus the QM view that accepts that quantum entanglement has no classical analogue. That you cannot analyse it in classical, realist terms.

It's not that I don’t know what change is, but I can think more abstractly about how a physically entangled system may be described. Retreating into the mathematics of states, perhaps, but thereby not being constrained by inappropriate classical thinking.
 
  • #110
AndreiB said:
Let's say the A measurement was UP. QM says that the state of B is DOWN (regardless if B was measured or not). If the A measurement did not change B and B is DOWN what was the state of B before the A measurement? The only answer is DOWN. If it was UP before the A measurement, and DOWN after the measurement it means that the A measurement did change B. If it was in an undecided state before the A measurement, and DOWN after the measurement it means that the A measurement did change B. If it was a 6-dimentional pink rabbit before the A measurement, and DOWN after the measurement it means that the A measurement did change B. And so on. The only state B could have had before the A measurement that is consistent with the requirement that it was not changed by the A measurement is DOWN. So, for that particular experiment we proved that the DOWN state of B is predetermined.
QM says that if A measured up, then B will measure down. The state of B before the A measurement was ##\hat{\rho}=1/2 \hat{1}##. The measurement at A is local, i.e., there is no faster-than-light influence at B's place. There's no way B can know that he will measure down before A has sent a message to him about the now certain outcome of his measurement. For both A and B what they observe on an ensemble of particles prepared in this entangled state are simply unpolarized particles. Only if they exchange their measurement protocols the 100% correlation is revealed.
 
  • #111
PeroK said:
This whole response simply emphasises the point that we cannot debate quantum entanglement because, put simply, you do not understand it.

That you believe you do simply compounds the problem.

In an entangled pair only the system of two particles has a state. The state of B cannot change, because B had no state before measurement. The measurement breaks the entanglement after which A and B have single-particle states.

It's clear you do not understand this or its implications for EPR.
That's not true. The subsystem has a state given by the partial trace over the subsystem A:
$$\hat{\rho}_B=\mathrm{Tr}_{A} \hat{\rho}.$$
It's called a reduced state.

If ##\hat{\rho}## is an entangled pure state, then ##\hat{\rho}## is a proper mixed state (i.e., not a pure state).
 
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  • #112
vanhees71 said:
If ##\hat{\rho}## is an entangled pure state, then ##\hat{\rho}## is a proper mixed state (i.e., not a pure state).
It's an improper mixed state.
 
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  • #113
vanhees71 said:
That's not true. The subsystem has a state given by the partial trace over the subsystem A:
$$\hat{\rho}_B=\mathrm{Tr}_{A} \hat{\rho}.$$
It's called a reduced state.

If ##\hat{\rho}## is an entangled pure state, then ##\hat{\rho}## is a proper mixed state (i.e., not a pure state).
Which does not change no matter what A does.
 
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  • #114
vanhees71 said:
QM says that if A measured up, then B will measure down.
Say A measured "UP". What is the state of B now, according to QM? Is it not "DOWN"?

vanhees71 said:
The state of B before the A measurement was ##\hat{\rho}=1/2 \hat{1}##.
Is this pre-measuremnt state the same as the post measurement one?

vanhees71 said:
The measurement at A is local, i.e., there is no faster-than-light influence at B's place.
I believe this to be true.

vanhees71 said:
There's no way B can know that he will measure down before A has sent a message to him about the now certain outcome of his measurement.
Who cares about what B knows? Let's say there is no B there, just a computer programmed to perform the measurement. All I care is about A.

vanhees71 said:
For both A and B what they observe on an ensemble of particles prepared in this entangled state are simply unpolarized particles. Only if they exchange their measurement protocols the 100% correlation is revealed.
So, let's say there is only one observer, A. He will only confirm the QM prediction when the measurement record from B arrives to him, at the speed of light. So what? He can look at the time of the B measurement and deduce that the measurements were space-like. Are you saying that he cannot rely on the experimental records?
 
  • #115
Demystifier said:
It's an improper mixed state.
I meant it's not a pure state, i.e., it's not of the form ##|\psi \rangle \langle \psi|##.
 
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  • #116
AndreiB said:
Say A measured "UP". What is the state of B now, according to QM? Is it not "DOWN"?
Well, that's again a matter of interpretation. I don't believe in the sense of the collapse postulate, and thus there's no change in the state outside of quantum theortical time evolution. To know what's the state after A's measurement I'd need to know how the measurement device is constructed and how to describe the interaction of A's photon with the measurement device (i.e., I'd need a Hamiltonian to model this interaction).
 
  • #119
After a Mentor discussion, thread will remain closed.
 

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